# How do you find the product of the complex number and its conjugate 1 + 3i?

Dec 24, 2015

You can use the identity $z \overline{z} = \text{Re"(z)^2+"Im} {\left(z\right)}^{2}$ to find:

$\left(1 + 3 i\right) \left(1 - 3 i\right) = {1}^{2} + {3}^{2} = 1 + 9 = 10$

#### Explanation:

Notice that:

$\left(a + b i\right) \left(a - b i\right) = {a}^{2} - {\left(b i\right)}^{2} = {a}^{2} - {i}^{2} {b}^{2} = {a}^{2} + {b}^{2}$

So we can deduce:

$z \overline{z} = \text{Re"(z)^2+"Im} {\left(z\right)}^{2}$

for any Complex number $z$